# Show that $(\sum a_{n}^{3} \sin n)$ converges given $\sum{a_n}$ converges

Given that $$\sum a_{n}$$ converges $$\left(a_{n}>0\right) ;$$ Then $$(\sum a_{n}^{3} \sin n)$$ is

My approach:

Since, $$\sum a_{n}$$ converges, we have $$\lim _{n \rightarrow \infty} n \cdot a_{n}$$ converges.

i.e. $$\left|n \cdot a_{n}\right| \leq 1$$ for $$n \geq K(\text { say })$$

$$\Rightarrow n \cdot a_{n}<1 \quad\left[\because a_{n}>0\right]$$

$$\Rightarrow a_{n}<\frac{1}{n}$$

$$\therefore a_{n}^{3}<\frac{1}{n^{3}}$$

$$\Rightarrow a_{n}^{3} \sin n \leq \frac{1}{n^{3}} \sin n \leq \frac{1}{n^{3}}$$

$$\Rightarrow \sum a_{n}^{3} \sin n \leq \sum \frac{1}{n^{3}}$$

$$\because \mathrm{RHS}$$ converges so LHS will also converge.

Any other better approach will be highly appreciated and correct me If I am wrong

• It is not true that if $\sum_{n=1}^{\infty} a_n$ converges then $a_n\leq 1/n$ for all sufficiently large $n$. You can have $a_n>1/n$ infinitely often. – Michael Jul 13 '20 at 19:34
• so my approach seems wrong – user791682 Jul 13 '20 at 19:34
• Your assertion Since, $\sum a_{n}$ converges, we have $\lim _{n \rightarrow \infty} n \cdot a_{n}$ converges. is wrong in general. It is true if $\{a_n\}$ is supposed to be positive decreasing. – mathcounterexamples.net Jul 13 '20 at 19:35
• You want to compare $\sum_{n=m}^{\infty} a_n^3 \sin(n)$ to $\sum_{n=m}^{\infty} a_n$. – Michael Jul 13 '20 at 19:36
• Is $a_n\ge 0$ an assumption? If not, take $a_n=\frac{\sin(n)}{n^{1/3}}$. Then, we have $\sum_n a_n$ converges. But does $\sum_n a_n^3\sin(n)=\sum_n \frac{\sin^4(n)}{n}$ converge? – Mark Viola Jul 13 '20 at 20:35

As $$\{a_n\}$$ is a positive sequence such that $$\sum a_n$$ converges, we have $$0 \le a_n \le 1$$ for $$n$$ large enough, say $$n \ge M$$.

Then for $$n \ge M$$

$$0 \le \vert a_n^3 \sin n \vert \le a_n^3 \le a_n$$

Hence $$\sum a_n^3 \sin n$$ converges absolutely.

Also a series $$\sum a_n$$ can be convergent while the sequence $$\{n a_n\}$$ diverges. Consider for example $$a_n$$ equals to $$0$$ if $$n$$ is not a square and equal to $$1/n$$ otherwise.

For $$a_n\geq 0$$ the sum converges as shown by @mathcounterexamaples.net

For general $$a_n$$ the statement is not true. The (deleted) counter-example given by @Mark Viola indeed is a counter-example. Taking $$a_n=\sin(n)/n^{1/3}$$ it suffices to show that $$\sum_{n\geq 1} a_n^3 \sin(n) =\sum_{n\geq 1} \frac{\sin^4 n}{n} =+\infty$$. To see this note that $$n$$ is asymptotically equidistributed mod $$2\pi$$ which implies that for any continuous function $$f\in C([-1,1])$$, $$\lim_{N\rightarrow +\infty} \frac{1}{N} \sum_{n=1}^N f(\sin(n)) = \frac{1}{2\pi} \int_0^{2\pi} f(\sin(t))\; dt$$. In particular, as $$N\rightarrow +\infty$$: $$M_N = \frac{1}{N} \sum_{n=1}^N \sin^4(n) \rightarrow \frac{1}{2\pi} \int_0^{2\pi} \sin^4(t)\; dt = \frac{3}{8}.$$ Proceeding as in Convergence of $\sum_n \frac{|\sin(n^2)|}{n}$ find a strictly increasing sequence $$N_k$$, $$k\geq 1$$ so that each $$M_{N_k}\geq \frac{1}{4}$$ and $$8 N_k \leq N_{k+1}$$. Then $$\sum_{n=1+N_k}^{N_{k+1}} \frac{\sin^4(n)}{n} \geq \frac{1}{N_{k+1}} \sum_{n=1+N_k}^{N_{k+1}} \sin^4(n) \geq M_{N_{k+1}} -\frac{N_k}{N_{k+1}} \geq \frac{1}{4} - \frac{1}{8}= \frac{1}{8}$$ Summing over $$k$$ implies the divergence of $$\sum_{n\geq 1} \frac{\sin^4(n)}{n}$$.

• Regarding showing $\sum_n \sin^4(n)/n$: an easier way is to write $8\sin^4(n)=3-4\cos(2x)+\cos(4x)$. Then $\sum_n \cos(a n)/n$ converges iff $a$ isn't a multiple of $2\pi$ and the harmonic series diverges. – FearfulSymmetry Jul 24 '20 at 19:21
• @Integrand Good point. – H. H. Rugh Jul 24 '20 at 20:04